UID: Jwb52z/Clay, M76106587 has a factor: 2253706211548544927116376383 (P-1, B1=690000, B2=12937500)

90.864 bits

P-1 found a factor in stage #2, B1=715000, B2=13227500.
UID: Jwb52z/Clay, M78208159 has a factor: 5980644118600514360767129 (P-1, B1=715000, B2=13227500)

82.307 bits

P-1 found a factor in stage #2, B1=690000, B2=12937500.
UID: Jwb52z/Clay, M76097309 has a factor: 160406892606660651307306068007 (P-1, B1=690000, B2=12937500)

97.018 bits.

I found a factor for the [URL="http://www.mersenne.ca/exponent/12973951"] exponent 12973951[/URL] which has 2^11 as part of the value for 'k'. Any exponent with a 'k' that includes a higher power than 11?

Hi tha,

you've asked for it, perhaps my highscore:[LIST][*][url]http://www.mersenne.ca/exponent/62419603[/url][/LIST]and some slightly lower powers of 2:[LIST][*][url]http://www.mersenne.ca/exponent/70213463[/url][*][url]http://www.mersenne.ca/exponent/60967087[/url][*][url]http://www.mersenne.ca/exponent/65490281[/url][*][url]http://www.mersenne.ca/exponent/72541463[/url][/LIST]
and some even lower powers of 3:[LIST][*][url]http://www.mersenne.ca/exponent/66150701[/url][*][url]http://www.mersenne.ca/exponent/59890121[/url][*][url]http://www.mersenne.ca/exponent/71694323[/url][/LIST]
and for 5:[LIST][*][url]http://www.mersenne.ca/exponent/68938921[/url][/LIST]
Oliver

If p is 1 (mod 4), then k can be 0 or 3 (mod 4), and if p is 3 (mod 4), then k can be 0 or 1 (mod 4). There is no k which is 2 (mod 4) because in this case 2kp+1 would be either 3 or 5 (mod 8) which is not possible for a factor.

Therefore, considering that 50% of the factors have a k which is 0 (mod 4), (i.e. 50% of the k's are multiple of 2^2 already) then about one in 1000 will have a k which is 0 (mod 2^11). As we have few millions of them...

Nice couple of factors for an exponent
 

[b]M78098261[/b] has a factor q=2715126139938490881846927116066757839; log2(q)=[b]121.03...[/b]
k = 11*13*3911*7321*81547*1802039*28890311 = 17382756704009650623635058379; log2(k)=93.812...

[b]M78098261[/b] has a factor q=56873105138451031651367; log2(q)=[b]75.59...[/b]
k = 19 * 23^2 * 29 * 163 * 7663739 = 364112493736903; log2(k)=48.371...

indeed not a every day factor, congratulations!

Oliver

P.S. I know there is nothing specials about these factors, small or big, B1-smooth or not, a factor is a factor. But bigger factors feel better than smaller ones (bigger is better!).

[QUOTE=TObject;403018][b]M78098261[/b] has a factor q=2715126139938490881846927116066757839; log2(q)=[b]121.03...[/b]
k = 11*13*3911*7321*81547*1802039*28890311 = 17382756704009650623635058379; log2(k)=93.812...

[b]M78098261[/b] has a factor q=56873105138451031651367; log2(q)=[b]75.59...[/b]
k = 19 * 23^2 * 29 * 163 * 7663739 = 364112493736903; log2(k)=48.371...[/QUOTE]

Which leads to the interesting thought, if we had run TF to 76 bits and found the factor, how long would it have been before we found the 121 bit one...

True.

Also, with slightly higher B1, the 76-bit factor could have shown up in the Stage 1 and the Stage 2 would not run at all.

P-1 found a factor in stage #2, B1=700000, B2=12950000.
UID: Jwb52z/Clay, M77149753 has a factor: 100126154465201540478799 (P-1, B1=700000, B2=12950000)

76.406 bits.